Question

In the given Fig., AC is diameter of the circle with centre O and A is point of contact, then find x.

Answer 1

GIVEN:
∠BAQ = 40°
OA ⟂ PQ
∠CAQ = 90°

[We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.]

∠CBA = 90°  [angle in a semicircle is a right angle]
∠CAQ = ∠CAB + ∠BAQ
90° =  ∠CAB + 40°
∠CAB = 90° - 40° = 50°

In ∆ABC,
∠CAB + ∠CBA + ∠BCA = 180°
[Angle sum property]
50° + 90° + x = 180°
140° + x = 180°
x = 180° - 140°

x = 40 °

Hence, the value of x is 40°.

HOPE THIS WILL HELP YOU..

Answer 2

Angle CBA=90degree (angle in semi circle )
AngleOAB+angleBAQ=90degree (radius in contact with tangent is perpendicular )
AngleOAB=90-40degree=50
by angle sum property
Angle CAB+angleB+angleC=10 degree
50+90+x=180
140+x=180
X=180-140=40degree